【leetcode】419. Battleships in a Board【E】

Given an 2D board, count how many different battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules:

• You receive a valid board, made of only battleships or empty slots.
• Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape 1xN (1 row, N columns) orNx1 (N rows, 1 column), where N can be of any size.
• At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.

Example:

X..X
...X
...X

In the above board there are 2 battleships.

Invalid Example:

...X
XXXX
...X

This is not a valid board - as battleships will always have a cell separating between them.

Your algorithm should not modify the value of the board.

Subscribe to see which companies asked this question

程序写的好长

不知道为啥这算是个E还过了这么多，估计没准儿自己程序写的不咋好。。。

分别横着竖着，如果一个船旁边儿有别的船，那就不是

class Solution(object):
    def countBattleships(self, board):

        bb = board
        m = len(board)
        n = len(board[0])

        b = [[u'.']*(n+2)]
        for i in bb:
            tmp = ([u'.'] + i + [u'.'])
            b += tmp[:],
        b += [u'.']*(n+2),


        i,j = 1,1
        res = 0
        while i < (m+1):
            j = 0
            while j < (n+1):
                flag = False
                while j < n+1 and b[i][j] == 'X':
                    if b[i-1][j] == 'X' or b[i+1][j] == 'X':
                        flag = False
                        break
                    b[i][j] = '.'
                    flag = True
                    j += 1
                if flag:
                    res += 1
                j += 1
            i += 1

        i = 1
        j = 1

        while j < n+1:
            i = 1
            while i < m+1:
                flag = False
                while i < m+1 and b[i][j] == 'X':
                    if b[i][j+1] == 'X' or b[i][j-1] == 'X':
                        flag = False
                        break
                    b[i][j] = '.'
                    flag = True
                    i += 1
                if flag:
                    res += 1

                i += 1
            j += 1

        return res